Question

Find the area of the given field. (all dimensions are in metre)

• A. 203 $m^2$
• B. 189 $m^2$
• C. 165 $m^2$
• D. 223 $m^2$

Answer 1

The correct option is D 223 $m^2$

Area of field =
Area of $△$DHC + Area of $△$DIE
+ Area of $△$AGF
+ Area of trapezium EIGF
+ Area of trapezium ABCH

Area of triangle = $\frac{1}{2}$$\times$base$\times$height

Area of $△$DHC = $\frac{1}{2}$$\times$HC$\times$DH
= $\frac{1}{2}$$\times$10$\times$(DI+IH) = $\frac{1}{2}$$\times$10$\times$(5+4)
= $\frac{1}{2}$$\times$10$\times$9 = 5$\times$9 = 45 m${}^2$

Area of $△$DIE = $\frac{1}{2}$$\times$IE$\times$DI
= $\frac{1}{2}$$\times$10$\times$5 = 5$\times$5 = 25 m${}^2$

Area of $△$AGF= $\frac{1}{2}$$\times$FG$\times$AG
= $\frac{1}{2}$$\times$4$\times$5 = 2$\times$5 = 10 m${}^2$

Area of trapezium
= $\frac{1}{2}$$\times$(sum of parallel sides)$\times$height

Area of trapezium EIGF
= $\frac{1}{2}$$\times$(EI+FG)$\times$(IG)
= $\frac{1}{2}$$\times$(10+4)$\times$(IH+HG)
= $\frac{1}{2}$$\times$14$\times$(4+5) = 7$\times$9
= 63 m${}^2$

Area of trapezium ABCH
= $\frac{1}{2}$$\times$(HC+AB)$\times$(AH)
= $\frac{1}{2}$$\times$(10+6)$\times$(GH+AG)
= $\frac{1}{2}$$\times$16$\times$(5+5) = $\frac{1}{2}$$\times$16$\times$10 = 8$\times$10
= 80 m${}^2$

Area of field = 45+25+10+63+80
= 223 $m^2$